Consider two long, straight, parallel wires separated by a distance of $d$. One wire carries a unifo

Question

Consider two long, straight, parallel wires separated by a distance of $d$. One wire carries a uniform linear charge density of $+2\lambda \mathrm{C} / \mathrm{m}$, and the other carries an equal and opposite uniform linear charge density of $-2\lambda \mathrm{C} / \mathrm{m}$. What is the magnitude of the net electric field at the mid-point of the line joining both the wires?

Professor PC Thomas Classes · Answer

The electric field due to an infinitely long straight wire with linear charge density $\lambda'$ at a perpendicular distance $r$ is given by the formula $E = \frac{\lambda'}{2 \pi \varepsilon_{0} r}$.

The point equidistant from both wires is the midpoint. The distance of this point from each wire is $r = d/2$. The magnitude of the linear charge density on each wire is $\lambda' = 2\lambda$.

The electric field from the positively charged wire, $\vec{E}_{+}$, points away from it. The electric field from the negatively charged wire, $\vec{E}_{-}$, points towards it. At the midpoint, both field vectors point in the same direction (from the positive wire towards the negative wire).

The magnitude of the electric field from each wire at the midpoint is:
$E_{+} = E_{-} = \frac{2\lambda}{2 \pi \varepsilon_{0} (d/2)} = \frac{2\lambda}{\pi \varepsilon_{0} d}$

Since both fields are in the same direction, the net electric field, $E_{net}$, is the sum of their magnitudes:
$E_{net} = E_{+} + E_{-} = \frac{2\lambda}{\pi \varepsilon_{0} d} + \frac{2\lambda}{\pi \varepsilon_{0} d} = \frac{4\lambda}{\pi \varepsilon_{0} d} \mathrm{~N} / \mathrm{C}$